# Proof including inverse trigonometric functions.

#### cdummie

Prove that:

arctan(x+y)<y+ arctanx y>0, x real number.

I don't know even how to start, usually with problems like this i use Rolle's or Lagrange's theorem but here, i can't prove it that way.

#### Archie

My first attempt would be to note that since $\arctan t$ is a strictly increasing function of $t$ we have that $\arctan a \gt \arctan b \implies a = \tan \arctan a \gt \tan \arctan b = b$.

Therefore we can write
$$\arctan (x+y) \gt y + \arctan x \iff x+y \gt \tan {(y + \arctan x)}$$
Then use $$\tan (A+B) = {\tan A + \tan B \over 1 - \tan A \tan B}$$

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